Let $h(x)=\begin{cases} \dfrac{6}{x+5}&\text{for }-5<x < -3 \\\\ x^2-6&\text{for }x\geq-3 \end{cases}$ Is $h$ continuous at $x=-3$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Solution: For $h$ to be continuous at $x=-3$, we need $\lim_{x\to -3}h(x)$ and $h(-3)$ to exist and be equal. Since $-3\geq -3$, the rule that applies to $x=-3$ is $x^2-6$. So $h(-3)=(-3)^2-6=3$. Now let's analyze $\lim_{x\to -3}h(x)$. Finding $\lim_{x\to -3^{ +}}h(x)$ For $x$ -values larger than $-3$, the appropriate rule for $h(x)$ is $x^2-6$. Since $x^2-6$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to -3^{ +}}h(x) \\\\ &=\lim_{x\to -3^{ +}}[x^2-6] \gray{x^2-6\text{ is the rule for }x>-3} \\\\ &=(-3)^2-6 \gray{x^2-6\text{ is continuous at }x=-3} \\\\ &=3 \end{aligned}$ Finding $\lim_{x\to -3^{ -}}h(x)$ For $x$ -values smaller than $-3$, the appropriate rule for $h(x)$ is $\dfrac{6}{x+5}$. Since $\dfrac{6}{x+5}$ is continuous for $-5<x < -3$, any limit in this interval is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to -3^{ -}}h(x) \\\\ &=\lim_{x\to -3^{ -}}\dfrac{6}{x+5} \gray{\dfrac{6}{x+5}\text{ is the rule for }x<-3} \\\\ &=\dfrac{6}{(-3)+5} \gray{\dfrac{6}{x+5}\text{ is continuous at }x=-3} \\\\ &=3 \end{aligned}$ Conclusion We found that: $\lim_{x\to -3^{ +}}h(x)=\lim_{x\to -3^{ -}}h(x)=h(-3)=3$ Since the one-sided limits are both equal to $h(-3)$, we can determine that the two-sided limit $\lim_{x\to -3}h(x)$ is also equal to $h(-3)$, and $h$ is continuous at $x=-3$.